# How can I find day of a certain month when the sum of all thursdays is a certain number?.

The problem is as follows:

The sum of all thursdays in certain month is $$80$$ and equal to the sum of all sundays in the following month, which has the same number of mondays than fridays it has this month. What day of the week is $$17th$$ of previous month from now?.

The alternatives given in my book are as follows:

$$\begin{array}{ll} 1.&\textrm{Sunday}\\ 2.&\textrm{Monday}\\ 3.&\textrm{Tuesday}\\ 4.&\textrm{Wednesday}\\ \end{array}$$

Does it exist a trick here?. I don't know exactly how can I get what it is being requested as it seems sort of convoluted. By attempting to do this by my own I think that in a month, regardless which month it is there are four weeks which you can select thursday, even february but only if it is from a leap year.

However if you select such february:

$$1+8+15+22+29=75$$

which doesn't seem the case as the result is off by $$80-75=5$$

hence it should be: (And it cannot be a february but march or any other month which has more than 30 days)

$$2+9+16+23+30=80$$

Since such month ends with thursday 30. But in order to also sum 80 it must have the exact arrangement of digits hence such month must be 31 days.

As Friday is 31st, Saturday is 1nd and sunday is 2nd.

Then such sum is also $$80$$.

This month can have either 30 days or 31 days.

Then the problem indicates that the current month has the same number of mondays and fridays as the month after it.

It can be seen that the current month starts on wednesday 1st, and goes on with thursday 2nd, friday 3rd, saturday 4th, sunday 5th, monday 6th, tuesday 7th, wednesday 8th, thursday 9th, friday 10th, saturday 11th and sunday 12th.

$$\textrm{Mondays= 6, 13, 20, 27 (only 4)}$$

$$\textrm{Fridays= 3, 7, 14, 28 (only 4)}$$

For the following month there will be the same number of days.

But since none of those exceeds $$30$$ days it can happen that the next month be $$30$$ days or $$31$$ days.

But the problem asks for which day of the week was $$17th$$ of the prior month from the current month and since the current month has $$31$$ days, the month before it could be $$31$$ days or $$30$$ days. As this happens between July and August or December and January, assuming the next month will have $$30$$ days.

This is the part where I'm in doubt. The month before now (from the perspective of the problem) would indicate that $$17th$$ (if such month has also $$31$$ days).

$$31-17=14$$ and it is divisible by seven.

Since we know the 1st of the current month is wednesday then 31st was tuesday, and $$17th$$ would be tuesday.

But if the month before it had $$30$$ days.

$$30-17=13$$ and it is not divisible by seven.

30th of the previous month would be tuesday and since it is behind by six days, $$17th$$ would be

$$\begin{array}{|c|c|c|c|c|c|c|}\hline W&T&F&S&S&M&T\\\hline 24&25&26&27&28&29&30\\\hline \end{array}$$

It would indicate that $$17th$$ would be wednesday.

And both alternatives appear. Which of them would be the answer?. Can someone help me here?.

As you have found, the certain month has Thursdays on $$2,9,16,23$$ and $$30$$. The following month has Sundays on those same dates. Since the second is a Sunday, the first is a Saturday. This means that the last day of the certain month is a Friday, and this must be the 31st because the 30th is a Thursday.
The following month has as many Mondays as the certain month has Fridays. We know these dates are $$3,10,17,24$$ and $$31$$, so both months have 31 days. The previous month therefore has 30 days. It begins on a Monday and ends on a Tuesday, and the 17th falls on Wednesday.